Let $A, B$ be $c^*$-algebras and $\phi:A\to B$ a positive, linear map. Then $\phi$ is bounded.
Proof: It is sufficient to proof boundedness of $\phi$ on the unitarization (I missunderstood that, see below..!) $A^+$ of A. Suppose $\phi$ is unbounded on $A^+$. Then there exists a sequence $(p_n)\subseteq A^+$ such that $\|p_n\|\le 1$ for all $n\in \mathbb{N}$ and $\|\phi(p_n)\|\ge n^3$. Let $p=\sum\limits_{n\in\mathbb{N}}\frac{p_n}{n^2}$, it is $\frac{p_n}{n^2}\le p$ and therefore $\|\phi(p)\|\ge \|\phi(\frac{p_n}{n^2})\|\ge n$ for every n, which is a contradiction.
I have some questions about the proof:
1.Why is it important that $(p_n)\subseteq A^+$ and not in $A$ in general? I don't see why this is important.
2.Why it is $\frac{p_n}{n^2}\le p$? the $p_n's$ (the summands) don't have to be positive I think.
3.I don't see the contradiction here. Why is $\|\phi(p)\|\ge n$ for every n impossible?
Regards
Edit: Ok, I missunderstood something, $A^+$ is the set of all positive elements of $A$ (see comments below), thank you martini. 1 and 2 are now clear.
Edit2: I see the contradiction now, $\|\phi(p)\|$ is constant, it couldn't be bigger as every n.
Because you are trying to prove (first) that $\phi$ is unbounded on $A^+$. You do this because you want to use that your map is positive, so it makes sense to work on the positive part of $A$.
The summands are positive.
Because $\|\phi(p)\|$ would be an upper bound for the natural numbers.